Three dice are tossed. What is the probability that the numbers shown will all be different?

Solution:

Given 3 fair dice, each of the dice has 6 faces.

So, the total number of outcomes will be 6 × 6 × 6 = 216

Now, if none of the numbers has to be repeated then

1. While rolling a die the possible outcomes are 6.

2. Now let’s assume the first die shows 2, then the second die should not show 2, then its possible outcomes are 5.

3. Similarly if the first die shows 2, and second die shows 3 then these two numbers should not be shown in the third die. Hence, its possible outcomes are 4.

So, the total number of favourable outcomes are calculated as 6 × 5× 4 = 120

Required probability P= favourable outcomes / total possible outcomes

P = 120/216

P = 5/9

Therefore, the probability that the numbers shown will all be different is 5/9.

Three dice are tossed. What is the probability that the numbers shown will all be different?

Summary :

If three dice are tossed, then the probability that the numbers shown will all be different is 5/9.